Tìm x, biết:
\(\sqrt{x^2-2x+2}+\sqrt{3x^3-6x+7}=-x^2+2x+2\)
Giải các phương trình dưới đây
1, \(\sqrt{9x^2-6x+2}+\sqrt{45x^2-30x+9}=\sqrt{6x-9x^2+8}\)
2,\(\sqrt{2x^2-4x+3}+\sqrt{3x^2-6x+7}=2-x^2+2x\)
3, \(\sqrt{6y-y^2-5}-\sqrt{x^2-6x+10}=1\) (x=3 ; y=3)
tìm x biết:
a)\(2x^2+5x+8+\sqrt{x}=x^2+3x+35+x^2+2x-7\)
b)\(3\sqrt{x}+7x+5=\sqrt{x}+4x-6+3x+18\)
c)\(8\sqrt{x}+2x-9=5x+7+6\sqrt{x}-3x-12\)
d)\(2\sqrt{3x}+11x-18=5x+3+6\sqrt{3x}+6x-21\)
Tìm tập xác định của hàm số :
a. y=\(\dfrac{1}{x^2-2x}+\sqrt{x^2-1}\)
b.y=\(\sqrt{x+1}+\sqrt{5-3x}\)
c.y=\(\sqrt{5x+3}+\dfrac{2x}{\sqrt{3-x}}\)
d.y=\(\dfrac{3x}{\sqrt{4-x^2}}+\sqrt{1+x}\)
e.y=\(\dfrac{5-2x}{(2-3x)\sqrt{1-6x}}\)
a: ĐKXĐ: x^2-2x<>0 và x^2-1>0
=>(x>1 và x<>2) hoặc x<-1
b: ĐKXĐ: x+1>0 và 5-3x>0
=>x>-1 và 3x<5
=>-1<x<5/3
c: DKXĐ: 5x+3>=0 và 3-x>0
=>x>=-3/5 và x<3
=>-3/5<=x<3
d: ĐKXĐ: 4-x^2>0 và 1+x>=0
=>x^2<4 và x>=-1
=>-2<x<2 và x>=-1
=>-1<=x<2
e: ĐKXĐ: 2-3x<>0 và 1-6x>0
=>x<>2/3 và x<1/6
=>x<1/6
6) \(\sqrt{x^2+12x+36}=-x-6\)
7) \(\sqrt{9x^2-12x+4}=3x-2\)
8) \(\sqrt{16-24x+9x^2}=2x-10\)
9) \(\sqrt{x^2-6x+9}==2x-3\)
10) \(\sqrt{x^2-3x+\dfrac{9}{4}}=\dfrac{3}{x}x-4\)
6) ĐKXĐ: \(x\le-6\)
\(\sqrt{\left(x+6\right)^2}=-x-6\Leftrightarrow\left|x+6\right|=-x-6\)
\(\Leftrightarrow x+6=x+6\left(đúng\forall x\right)\)
Vậy \(x\le-6\)
7) ĐKXĐ: \(x\ge\dfrac{2}{3}\)
\(pt\Leftrightarrow\sqrt{\left(3x-2\right)^2}=3x-2\Leftrightarrow\left|3x-2\right|=3x-2\)
\(\Leftrightarrow3x-2=3x-2\left(đúng\forall x\right)\)
Vậy \(x\ge\dfrac{2}{3}\)
8) ĐKXĐ: \(x\ge5\)
\(pt\Leftrightarrow\sqrt{\left(4-3x\right)^2}=2x-10\)\(\Leftrightarrow\left|4-3x\right|=2x-10\)
\(\Leftrightarrow4-3x=10-2x\Leftrightarrow x=-6\left(ktm\right)\Leftrightarrow S=\varnothing\)
9) ĐKXĐ: \(x\ge\dfrac{3}{2}\)
\(pt\Leftrightarrow\sqrt{\left(x-3\right)^2}=2x-3\Leftrightarrow\left|x-3\right|=2x-3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=2x-3\left(x\ge3\right)\\x-3=3-2x\left(\dfrac{3}{2}\le x< 3\right)\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=2\left(tm\right)\end{matrix}\right.\)
gpt:\(\sqrt{3x^2+6x+4}+\sqrt{2x^2+4x+11}=\left(1-x\right)\left(x+3\right)\)
\(\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+21}=5-x^2-2x\)
\(\sqrt{x^2-x+2}+\sqrt{x^2-3x+6}=2x\)
Tìm x:
a. \(\sqrt{9x^2}=2x+1\)
b. \(\sqrt{x^2+6x+9}=3x-1\)
c. \(\sqrt{x^2-2x+4}=2x-3\)
\(a,\sqrt{9x^2}=2x+1\\ \Leftrightarrow\left[{}\begin{matrix}3x=2x+1,\forall x\ge0\\-3x=2x+1,\forall x< 0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1,\forall x\ge0\left(N\right)\\x=-1,\forall x< 0\left(N\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
\(b,\sqrt{x^2+6x+9}=3x-1\\ \Leftrightarrow\sqrt{\left(x+3\right)^2}=3x-1\\ \Leftrightarrow\left[{}\begin{matrix}x+3=3x-1,\forall x+3\ge0\\x+3=1-3x,\forall x+3< 0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2,\forall x\ge-3\left(N\right)\\x=-\dfrac{1}{2},\forall x< -3\left(L\right)\end{matrix}\right.\Leftrightarrow x=2\)
\(c,\sqrt{x^2-2x+4}=2x-3\left(x\in R\right)\\ \Leftrightarrow x^2-2x+4=\left(2x-3\right)^2\\ \Leftrightarrow x^2-2x+4=4x^2-12x+9\\ \Leftrightarrow3x^2-10x+5=0\\ \Delta=100-4\cdot3\cdot5=40\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{10-\sqrt{40}}{6}\\x=\dfrac{10+\sqrt{40}}{6}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5-\sqrt{10}}{3}\\x=\dfrac{5+\sqrt{10}}{3}\end{matrix}\right.\)
\(a.\sqrt{9x^2}=2x+1\)
<=> \(\sqrt{9}x=2x+1\)
<=> 3x = 2x + 1
<=> 3x - 2x = 1
<=> x = 1
Tìm x:
\(\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+21}=5-x^2-2x\)
Ta có: \(VT=\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+21}\)
\(=\sqrt{3x^2+6x+3+4}+\sqrt{5x^2+10x+5+16}\)
\(=\sqrt{3\left(x+1\right)^2+4}+\sqrt{5\left(x+1\right)^2+16}\ge2+4=6\)
Ta có: \(VP=5-x^2-2x\)
\(=-\left(x^2+2x+1\right)+6\)
\(=-\left(x+1\right)^2+6\le6\)
VP=VT khi x+1=0
hay x=-1
Vậy: x=-1
Tìm x:
\(\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+14}=4-2x-x^2\)
\(VT=\sqrt{3\left(x+1\right)^2+4}+\sqrt{5\left(x+1\right)^2+9}\ge\sqrt{4}+\sqrt{9}=5\)
\(VP=5-\left(x+1\right)^2\le5\)
Đẳng thức xảy ra khi và chỉ khi:
\(\left(x+1\right)^2=0\Leftrightarrow x=-1\)
Giải pt:
a)\(\sqrt{x^2+3x+2}+\sqrt{x^2+6x+5}=\sqrt{2x^2+9x+7}\)
b)\(\sqrt{2x+3}+\sqrt{x+1}=3x+2\sqrt{2x^2+5x+3}-2\)
a/ ĐKXĐ: \(\left[{}\begin{matrix}x\ge-1\\x\le-5\end{matrix}\right.\)
Bình phương 2 vế:
\(x^2+3x+2+2\sqrt{\left(x^2+3x+2\right)\left(x^2+6x+5\right)}+x^2+6x+5=2x^2+9x+7\)
\(\Leftrightarrow2\sqrt{\left(x^2+3x+2\right)\left(x^2+6x+5\right)}=0\)
\(\Rightarrow\left[{}\begin{matrix}x^2+3x+2=0\\x^2+6x+5=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-1\\x=-2\left(l\right)\\x=-5\end{matrix}\right.\)
Vậy pt có 2 nghiệm \(x=-1;x=-5\)
b/ ĐKXĐ: \(x\ge-1\)
Đặt \(\sqrt{2x+3}+\sqrt{x+1}=a>0\Rightarrow a^2-6=3x+2\sqrt{2x^2+5x+3}-2\)
Phương trình trở thành:
\(a=a^2-6\Leftrightarrow a^2-a-6=0\Rightarrow\left[{}\begin{matrix}a=-2\left(l\right)\\a=3\end{matrix}\right.\)
\(\Rightarrow\sqrt{2x+3}+\sqrt{x+1}=3\Leftrightarrow3x+4+2\sqrt{2x^2+5x+3}=9\)
\(\Leftrightarrow2\sqrt{2x^2+5x+3}=5-3x\)
\(\Leftrightarrow\left\{{}\begin{matrix}5-3x\ge0\\4\left(2x^2+5x+3\right)=\left(5-3x\right)^2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{5}{3}\\x^2-50x+13=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=25+6\sqrt{17}\left(l\right)\\x=25-6\sqrt{17}\end{matrix}\right.\)
Vậy pt có nghiệm duy nhất \(x=25-6\sqrt{17}\)
a) \(\sqrt{\left(x+1\right)\left(x+2\right)}+\sqrt{\left(x+1\right)\left(x+5\right)}=\sqrt{\left(x+1\right)\left(2x+7\right)}\)
\(ĐK\Leftrightarrow\left[{}\begin{matrix}x\le-1\\x\ge-2\end{matrix}\right.\)
\(\Leftrightarrow\sqrt{\left(x+1\right)\left(x+2\right)}+\sqrt{\left(x+1\right)\left(x+5\right)}-\sqrt{\left(x+1\right)\left(2x+7\right)}=0\)
\(\Leftrightarrow\sqrt{\left(x+1\right)}\left(\sqrt{x+2}+\sqrt{x+5}-\sqrt{2x+7}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\\sqrt{x+2}+\sqrt{x+5}=\sqrt{2x+7}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x+2+x+5+2\sqrt{\left(x+2\right)\left(x+5\right)}=2x+7\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\2\sqrt{\left(x+2\right)\left(x+5\right)}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-2\\x=-5\end{matrix}\right.\)
vậy \(S=\left\{-1;-2;-5\right\}\)
b) \(\sqrt{2x+3}+\sqrt{x+1}=3x+2\sqrt{\left(2x+3\right)\left(x+1\right)}-2\)
ĐK \(\left[{}\begin{matrix}x\ge-1\\x\le-\dfrac{3}{2}\end{matrix}\right.\)
\(\Leftrightarrow\sqrt{2x+3}+\sqrt{x+1}=2x+3+2\sqrt{\left(2x+3\right)\left(x+1\right)}+x+1-6\)
\(\Leftrightarrow\left(\sqrt{x+1}+\sqrt{2x+3}\right)^2-\left(\sqrt{x+1}+\sqrt{2x+3}\right)-6=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}+\sqrt{2x+3}=3\\\sqrt{x+1}+\sqrt{2x+3}=-2\left(VL\right)\end{matrix}\right.\)
\(\Leftrightarrow\left(\sqrt{x+1}+\sqrt{2x+3}\right)^2=9\)
\(\Leftrightarrow x+1+2\sqrt{\left(x+1\right)\left(2x+3\right)}+2x+3=9\)
\(\Leftrightarrow2\sqrt{\left(x+1\right)\left(2x+3\right)}=5-3x\)
\(\Leftrightarrow4\left(2x^2+5x+3\right)=\left(5-3x\right)^2\) (đk \(x\le\dfrac{5}{3}\) )
\(\Leftrightarrow8x^2+20x+12=9x^2-30x+25\)
\(\Leftrightarrow x^2-50x+13=0\)
bấm máy rồi so với đk ra nghiệm nha bạn